Shrinking numbers

Question

The input is an array of (at least 3, maximum 20) different integers. Each integer is greater than -1000 and smaller than 1000.

Your task is to shrink the numbers by "linearly mapping" them from 0.0 to 1.0. This means the smallest number in the array will be mapped to 0.0, the largest to 1.0.

You get the array as a parameter (inside a function) or stdin/program arguments (you can choose). Print out the result in the format double1;double2;double3;.... The output has to have the same order as the input.

If you want, you can round the output to 2 digits after the decimal point. There must be atleast 1 digit after the decimal point.

The usage of built-in functions (functions which scale down the numbers for you, such as mathematicas Rescale) is disallowed.

Examples:

Input              Output
[5,-20,30]         0.5;0.0;1.0
[1,2,3,4,5]        0.0;0.25;0.5;0.75;1.0
[0,5,100,400]      0.0;0.01;0.25;1.0

(The last output is rounded, otherwise it would be 0.0;0.0125;0.25;1.0)

Answer 1

CJam, 18 bytes

q~_$0=f-_$W=df/';*

Note that the online interpreter wrongfully represents 0d as 0 instead of 0.0.

Example run

$ cjam shrink.cjam <<< '[5 -20 30]'; echo
0.5;0.0;1.0
$ cjam shrink.cjam <<< '[1 2 3 4 5]'; echo
0.0;0.25;0.5;0.75;1.0
$ cjam shrink.cjam <<< '[0 5 100 400]'; echo
0.0;0.0125;0.25;1.0

How it works

q~                    " P := eval(input())         ";
  _$0=                " S := sorted(P)[0]          ";
      f-              " Q := { X - S : X ∊ P }     ";
        _$W=d         " D := double(sorted(Q)[-1]) ";
             f/       " R := { X / D : X ∊ Q }     ";
               ';*    " print(join(R, ';'))        ";

Answer 2

JavaScript, ES6, 81 bytes

Thanks to @edc65 for the toFixed trick

F=a=>a.map(v=>((v-n)/d).toFixed(2),n=Math.min(...a),d=Math.max(...a)-n).join(';')

Run it in latest Firefox Console.

This creates a function f which you can invoke like

F([5,-20,30])

Answer 3

Python 2, 72 68 63 56 55

Obviously not as concise as other answers, but anyway:

x=input()
m=min(x)
print[(i*1.-m)/(max(x)-m)for i in x]

Sample run:

[1,100,25,8,0]                  #input
[0.01, 1.0, 0.25, 0.08, 0.0]    #output

---

Old (68 characters, written in Python 3):

x=eval(input())
y=sorted(x)
print([(i-y[0])/(y[-1]-y[0])for i in x])

Answer 4

CJam, 24 23 bytes

l~_$)\(:M\;-\Mf-\df/';*

Input be like:

[5 -20 30]

Try it online here Note that online compiler prints Double 0 as 0 only. Run in the java interpreter which prints correctly.

How it works:

l~                      "Evaluate input and convert each element to double";
  _$                    "Copy the array and sort the copied array";
    )                   "Pop the last element out of the array. This is Max";
     \                  "Swap last two stack elements, bring sorted array on top";
      (:M               "Pop the first element of array and store it in M. This is Min";
         \;             "Bring the remaining of sorted array on top and remove it from stack";
           -\           "Subtract Max and Min and bring the original array to top of stack"
             Mf-        "Push min to stack and subtract it from each array element";
                \df/    "Bring (Double)(Max-Min) to top and divide each array element by it";
                   ';*  "Push the character ; to stack and join the array with it";

Answer 5

C# 92

Running inside LinqPad

void F(int[]a)
{
   double n=a.Min(),d=a.Max()-n;
   a.Select(x=>((x-n)/d).ToString("0.00")).Dump();
}

Test in LinqPad

void Main()
{
    F(new int[]{5,-20,30});
}
void F(int[]a){double n=a.Min(),d=a.Max()-n;a.Select(x=> ((x-n)/d).ToString("0.00")).Dump();}

Output

IEnumerable<String> (3 items)
0,50 
0,00 
1,00 

Answer 6

APL (15)

(2⍕+÷⌈/)(+-⌊/)⎕

(or, without trains, also 15 characters:)

2⍕V÷⌈/V←V-⌊/V←⎕

This reads the argument from the keyboard and prints the result to the screen.

Explanation:

• ⎕: read a line from the keyboard and evaluate it
• +-⌊/: subtract the lowest item in the array from all items in the array
• +÷⌈/: divide each item in the array by the array's highest item
• 2⍕: format with two decimal places

Test:

      (2⍕+÷⌈/)(+-⌊/)⎕
⎕:
     5 ¯20 30
 0.50 0.00 1.00
      (2⍕+÷⌈/)(+-⌊/)⎕
⎕:
      1 2 3 4 5
 0.00 0.25 0.50 0.75 1.00
      (2⍕+÷⌈/)(+-⌊/)⎕
⎕:
      0 5 100 400
 0.00 0.01 0.25 1.00

Answer 7

Pyth, 18

Now with correct formatting!

j\;mc-dhSQ-eSQhSQQ

Test:

$ pyth -c 'j\;mc-dhSQ-eSQhSQQ' <<< '[0,5,100,400]'
0.0;0.0125;0.25;1.0

Explanation:

(implicit)              Q = eval(input())
j\;                     ';'.join(
   m                             map(lambda d:
    c                                         float_div(
     -dhSQ                                              d-sorted(Q)[0],
     -eSQhSQ                                            sorted(Q)[-1]-sorted(Q)[0]),
    Q                                         Q))

Answer 8

Octave 25

b=min(l);(l-b)/(max(l)-b)

Assumes input is in l and since it's an interactive shell the result is printed automatically (is this allowed?)

Answer 9

APL, 31 characters / 55 bytes

{b←⌊/⍵⋄c←(⌈/⍵)-b⋄{2⍕(⍵-b)÷c}¨⍵}

Old code without digits after decimal point:

{b←⌊/⍵⋄c←(⌈/⍵)-b⋄{(⍵-b)÷c}¨⍵}

Take minimum of vector, take difference between maximum and minimum of vector, subtract the minimum from each element and divide by the difference between min and max.

Edited Code to print two digits after the decimal point:

Answer 10

CJam, 30 29 bytes

l~:d_{e>}*\_{e<}*:Mf-\M-f/';*

Expects the input on STDIN like [5 -20 30].

Test it here. (This will print integer 0 and 1 without decimal point, but the Java interpreter does print 0.0 and 1.0.)

Due to a bug I can't shorten {e>}* to :e> although that should be possible as per the spec (which would save 4 bytes when applied to both min and max).

Slightly outdated explanation: (will amend later)

l~:d_{e<}*_@_{e>}*@-\@f-\f/';* "Read and eval the input leaving an array of strings on the stack";
l~                             "Read and eval the input leaving an array of strings on the stack";
  :d                           "Convert all elements to double";
    _                          "Duplicate the array";
     {e<}*                     "Wrap the MIN function in a black and fold it onto the array";
          _                    "Duplicate the minimum";
           @                   "Rotate the stack, pulling the array to the top";
            _                  "Duplicate the array";
             {e>}*             "Same as before, now with MAX";
                  @            "Rotate the stack, pulling the minimum to the top";
                   -           "Subtract to give the total range";
                    \          "Swap range and array";
                     @         "Rotate the stack, pulling the other minimum to the top";
                      f-       "Subtract the minimum from each element in the array";
                        \      "Swap range and array";
                         f/    "Divide each element in the array by the range";
                           ';  "Push a semicolon character";
                             * "Riffle the semicolon into the array";

At the end of the program, the stack contents are printed by default.

I'm sure there is a way to save half of the stack reshuffling, but I'm not that comfortable with CJam yet.

Answer 11

Xojo, 179 bytes

dim x,n as double,k,z as int16,s() as string
n=1e3
x=-n
for each k in a
x=max(x,k)
n=min(n,k)
next
for k=0 to ubound(a)
s.append str((a(k)-n)/(x-n),"0.0#")
next
msgbox join(s,";")

Answer 12

R, 60 bytes

m=min(x<-scan());cat(sprintf("%f",(x-m)/(max(x)-m)),sep=";")    

Formatting eats up a lot of bytes due to 0 and 1 by default are trimmed to display nothing past the integer part.

Answer 13

Clojure 63

(fn[v](let[l(apply min v)](map #(/(- % l)(-(apply max v)l))v))) 

Doesn't quite follow the rules, since it returns fractions instead of doubles. If that's not acceptable, add 7 bytes

Ungolfed:

(fn [values]
    (let [low (apply min values)]
         (map #(/ (- % low)
                  (- (apply max values) low))
              values)))

Callable like this:

((fn[v](let[l(apply min v)](map #(/(- % l)(-(apply max v)l))v))) [5 -20 30])

Output: (1/2 0 1)

Answer 14

Ruby, 49

f=->a{$><<a.map{|x|(x-l=a.min).fdiv(a.max-l)}*?;}

Explanation:

f=->a{}     # Define a lambda that takes one argument a
$><<        # Print the following to STDOUT
a.map{|x|}  # For each element x
(x-l=a.min) # Find the lowest element of a, assign it to l, and subtract it from x
.fdiv       # Float division (/ truncates)
(a.max - l) # Divide by the maximum minus the minimum
*?;         # Convert the resulting array into a string joined by the ';' character

Answer 15

05AB1E, 13 bytes (Non-competing)

ZsWD1.S*+s/Z/

Try it online!

Answer 16

Q (31) IMPROPER OUTPUT FORMAT

{(%/)(x;max x)-min x}(.:)(0::)0

input

1 2 3

output

0 .5 1

Answer 17

Perl - 60

my@a=sort@ARGV;print map{($_-$a[0])/($a[-1]-$a[0])." "}@ARGV

Answer 18

Java 7, 149 bytes

float[]c(int[]x){int b=1<<31,a=b-1,j=0,l=x.length;for(int i:x){a=i<a?i:a;b=i>b?i:b;}float[]r=new float[l];for(;j<l;r[j]=x[j++]-a)*1f/(b-a);return r;}

Ungolfed & test code:

Try it here.

import java.util.Arrays;
class M{
  static float[] c(int[] x){
    int b = Integer.MIN_VALUE,
        a = b-1, // In Java, Integer.MIN_VALUE - 1 = Integer.MAX_VALUE (and vice-versa)
        j = 0,
        l = x.length;
    for(int i : x){
      a = i < a ? i : a; // Determine min value of array
      b = i > b ? i : b; // Determine max value of array
    }
    float[] r = new float[l];
    for(; j < l; r[j] = (x[j++] - a) * 1f / (b-a));
    return r;
  }

  public static void main(String[] a){
    System.out.println(Arrays.toString(c(new int[]{ 5, -20, 30 })));
    System.out.println(Arrays.toString(c(new int[]{ 1, 2, 3, 4, 5 })));
    System.out.println(Arrays.toString(c(new int[]{ 0, 5, 100, 400 })));
  }
}

Output:

[0.5, 0.0, 1.0]
[0.0, 0.25, 0.5, 0.75, 1.0]
[0.0, 0.0125, 0.25, 1.0]

Answer 19

Vyxal, 5 bytes

g-:G/

Try it Online!

Outputs as arbitrary-precision rationals. Look Ma, no Unicode!

g-    # Subtract the minimum
  :G  # Make a copy and get the maximum of that
    / # Divide (list - min) by the maximum

Answer 20

Ly, 33 bytes

&n&sr>las<l/uy[';ofl/0L[!f!]pu,];

Try it online!

This is pretty direct translation of the requirements, but there are two things which add code I wanted to avoid... First, since 0 is a valid number in the list, the code has to explicitly add (and manipulate) a loop iterator on the stack. And second, since the rules imply that negative numbers should produce a 0 output, there's some extra code for that. If -0.0 is an acceptable output, the code could be a few bytes shorter. :)

&n&sr                             (1) read input and prep for processing
&n     - read STDIN onto the stack
  &s   - save a copy of the stack 
    r  - reverse the stack

     >las<                        (2) find/stash the largest input number
>       - switch to an empty stack
 l      - restore a copy of the input list
  a     - sort the stack
   s    - save the top of stack (largest number) to backup cell
    <   - switch back to the stack with the original input

          l/uy                    (3) print first entry, prep for loop
 l     - load the largest number from the backup cell     
  /    - divide to get the 0-1 scaled number
   u   - print as a number
    y  - push the stack size onto the stack  

              [';ofl/0L[!f!]pu,]; (4) process 2-N list entries
 [                ]   - loop through all the list entries
  ';o                 - print the ";" field separator
     f                - flip the iterator and next list entry on stack
      l               - load the largest number on the input
       /              - divide to get the 0-1 ratio
        0L            - compare result against 0, looking for "<0"
          [   ]p      - if/then structure, true if result <0
           !f!        - true: make top two stack entries "0"
                u     - print ratio (or "0" if result was <0)
                 ,    - decrement iterator
                   ;  - exit program to avoid print extraneous "0"

If -0.0 is an acceptable output, this 4 bytes shorter version could be used.

 &n&sr>las<l/uy[';ofl/0L!*u,];